Question

2. Probability (30%). Figure out the probability in the following scenarios.

(a) A number generator is able to generate an integer in the range of [1, 100], where each number has equal chances to be generated. What is the probability that a randomly generated number x is divisible by either 2 or 3, i.e., P(2 | x or 3 | x)? (5%)

(b) In a course exam, there are 10 single-choice questions, each worthing 10 points and having 4 choices (A, B, C, and D, with only one correct). There is one student, denoted as s, who has learned nothing from the course and hence has to randomly guess the answers. That means for any question, each one of the four choices has equal chances to be picked up by s. What is the probability that s passes the exam (earning a total of ≥ 60 points)? (5%)

(c) Consider three positive integers, x1, x2, x3, which satisfy the inequality below: x1 + x2 + x3 = 17. (1) Let’s assume each element in the sample space (consisting of solution vectors (x1, x2, x3) satisfying the above conditions) is equally likely to occur. For example, we have equal chances to have (x1, x2, x3) = (1, 1, 15) or (x1, x2, x3) = (1, 2, 14). What is the probability the events x1 + x2 ≤ 8 occurs, i.e., P(x1 + x2 ≤ 8 | x1 + x2 + x3 = 17 and x1, x2, x3 ∈ Z+) (Z+ is the set containing all the possible positive integers)? (5%)

(d) There are unlimited fake coins and only one real coin. The fake coins and the real coin are almost the same and can only be detected by a special machine. At the very beginning, there are two coins in a bag, one fake and the other real (but we don’t know which one is real). We continue the following process till the real coin is found: At the each step, we randomly sample one coin from the bag and examine whether it is fake. If yes, we put the coin back to the bag, additionally put in another fake coin, and randomly draw a coin for examination. The sampling process won’t stop until we find the real coin. Assuming that each coin (either fake or real) has equal chances to be selected, what is the probability that we sample 9 times but still cannot find the real coin (and hence has to continue the sampling process)? (5%)

(e) From a random sports news, the probability of observing the word “ball” and “player” is 0.8 and 0.7, respectively. For a non-sports news, the probability to observe “ball” is 0.1, so does that to observe “player”. Let’s assume that in any article, the appearance of any two words (including “ball” and “player”) are independent with each other. Also, the probability of sports news’ occurrences is 0.2. Given a news report x containing both “ball” and “player”, what is the probability that x is a sports news. (10%)

Answer 1

a) Required probability = (No. of favorable choices)/Sample Space

Sample space = 100

No. of favorable choices = Total number of numbers divisible by either 2 or 3

Mathematically

  

Hence, the required probability = (answer)

b) Since the possibility of choosing any option is equally likely, and since there is only one option correct

Hence, the probability of choosing the correct option = 1/4 = 0.25

And the probability of choosing the wrong option = 1-0.25 = 0.75

We will use Bernoulli's distribution here to calculate the required probability

where p = 0.25

Now, of the 10 questions, at least 6 must be correct to pass the exam.

The Bernoulli's distribution

Where n is the total number of questions, x is the number of correct answers and p is the probability for a correct answer.

For x = 6

For x = 7

For x = 8

For x = 9

For x = 10

Hence, on adding the all above probabilities, we get the total probability of scoring equal to or more than 60points in the test = 0.016+0.003+0.0004+0.00003+0.00000 = 0.019 (answer)

c) The equation is

or,

We need to find the probability of

Since are both positive integers, hence must be greater than or equal to 2.

Hence, finding the probability of is same as finding the probability for

No. of favorable values for

Now, The maximum value for will be possible when will be minimum.

Hence, for and ,   = 17 - 1 -1 = 15

Hence maximum value for = 15

And, minimum value for = 1

Total number of possible values for = 15

Required probability of   

= (No. of favorable values for )/(No. of possible values for )  

(answer)

d) In this question, we need to find the probability of not selecting a real coin in the first 9 attempts

Probability of selecting a real coin = 0.5 and of selecting a fake coin = 0.5

Since the number of coins in the bag continues to increase after every unsuccessful attempt. hence the probability of selecting a fake coin will keep on increasing

Now, the required probability will be equal to the product of probabilities of each attempt till the 9th attempt

(answer)

e) The fifth question is a question of conditional probability

Let us define the events first

A = X is a sports news

B = The report X contains both ball and player

We need to find the Conditional probability P(A / B) that is, Probability of x being a sports news given that the news contains both the word 'ball' and 'player'.

The formula for conditional probability

Now,

Hence, there is a probability of 0.93 that if a news contains the words 'ball' and 'player' then the news will be a sports news.